Biomedical Engineering Reference
In-Depth Information
z
1
tan
1
where c ¼ p þ
p
. It is always helpful to check analysis results and one easy point to
z
2
check is usually at time zero
.
"
#
"
p
1
1
#
z
2
g
K
cos
ðfÞ
1
g
K
þ
yð
0
Þ¼
1
þ
p
¼
1
p
¼
0
z
2
z
2
as it should, since the saccade starts at primary position or y(0)
¼
0.
To fully explore the quality of a model, it is necessary to compare its performance against
the data. For a saccade, convenient metrics are time to peak overshoot, which gives an indi-
cation of saccade duration and peak velocity. These metrics were discussed previously
when the main sequence diagram was described in Section 13.2.
The time to peak overshoot of saccade model,
T
p
, is found by first calculating
"
!
#
t
¼
T
p
¼
q
1
e
zo
n
t
1
@
@
t
¼
@
y
g
K
z
2
1
þ
p
cos o
n
t
þ
c
0
ð
13
:
6
Þ
@
t
z
2
Using the chain rule to evaluate Eq. (13.6) and substituting
t
¼
T
p
, yields
p
o
d
¼
p
p
1
T
p
¼
:
ð
13
:
7
Þ
z
2
o
n
p
p
With Westheimer's constants of z
¼
0
:
7 and o
n
¼
120, we find that
T
p
¼
¼
37 ms
:
7
2
120
1
0
:
T
p
in the Westheimer model is independent of saccade size and not in agreement with experi-
mental data presented in the main sequence diagram. The data show a saccade duration that
increases with saccade amplitude, where this model has a constant duration.
EXAMPLE PROBLEM 13.2
Show that Eq. (13.7) follows from Eq. (13.6). Find the value of y(
T
p
).
Solution
From Eq. (13.6), we have
g
K
h
i
e
zo
n
t
o
d
e
zo
n
t
sin o
d
T
p
þ c
p
zo
n
cos o
d
T
p
þ c
¼
0
ð
13
:
8
Þ
z
2
1
Eq. (13.8) is rewritten as
q
1
¼
z
2
z cos o
d
T
p
þ c
sin o
d
T
p
þ c
which yields
¼
z
1
tan o
d
T
p
þ
c
p
¼
tan c
ð
13
:
9
Þ
z
2
